Problem: $f(x) = \dfrac{ \sqrt{ 16 - x } }{ \sqrt{ x - 9 } }$ What is the domain of the real-valued function $f(x)$ ?
Solution: First, we need to consider that $f(x)$ is undefined anywhere where either radical is undefined, so anywhere where either radicand (the expression under the radical symbol) is less than zero. The top radical is undefined when $16 - x < 0$ So the top radical is undefined when $x > 16$ , so we know $x \leq 16$ The bottom radical is undefined when $x - 9 < 0$ So the bottom radical is undefined when $x < 9$ , so we know $x \geq 9$ Next, we need to consider that $f(x)$ is undefined anywhere where the denominator, $\sqrt{ x - 9 }$ , is zero. So $\sqrt{ x - 9 } \neq 0$ , so $x - 9 \neq 0$ , so $x \neq 9$ So we have three restrictions: $x \leq 16$ $x \geq 9$ , and $x \neq 9$ Combining these three restrictions, we know that $9 < x \leq 16$ Expressing this mathematically, the domain is $\{ \, x \in \RR \mid 9< x \leq16\, \}$.